a hot body is placed in cooler surroundings.when the body temperature is 75 degree celcious the rate of cooling is 4degree celcius/min when it is 55 degree celiciusthe rate of cooling is 2 degree celcious/min. the temperature of the surroundings is

∆t1-∆t2/T=(∆t1+∆t2/2)-t° T-time, t°-temperature of surroundings As the rate of cooling of hot body is 4°c/min(75°c) ∆t1-∆t2/T=(∆t1+∆t2/2)-t° 75-71/60=(75+71/2)-t° 4/60=146-2t°/2 — (equation no-1) For 55°c rate of cooling is 2°c/min ∆t1-∆t2/T=(∆t1+∆t2/2)-t° 55-53/60=(55+53/2)-t° 2/60=108-2t°/2 —(equation no-2) Dividing eq no 1 with eq no 2 4/2=146-2t°/108-2t° 2=146-2t°/108-2t° 2(108-2t°)=146-2t° 216-4t°=146-2t° 216-146=4t°-2t° 70=2t° t°=35°c